Gauss Law Formula

Gauss Law is just a law that is general to any closed area that permits the calculation of the field of an enclosed charge by mapping the field for a surface out of the charge distribution. It simplifies the analysis associated with the geometrical symmetry shape of the eclectic field.

What is the Gauss Law?

The total electric flux out of a closed surface is equal to the charge enclosed upon the permittivity of the material. Gauss Law is just a fundamental law applicable to any area that is closed. It’s a crucial tool since it allows the evaluation associated with the amount of enclosed electric charge. For geometries having adequate symmetry, it simplifies the calculation for the electric field.

Significance of Gauss Law

This is to explain the significance of Gauss law. Gauss’s Law statement is correct and ideal for any closed surface in addition to the size or shape of the object.

The word Q within the formula of gauss law indicates the summation of all of the electric charges, which are enclosed within the object regardless of the position of the charge on the surface.

The selected surface for the functionality of gauss law is known as the Gaussian area or surface. Still, this area should not be passed away through any kind or variety of isolated charges.

This is mainly used by the simplified evaluation regarding the electrostatic scenario where the system holds some equilibrium, which can only occur whenever we choose a closed surface.

The formula of Gauss Law

According to the Gauss theorem, the total electric charge enclosed within a closed surface is proportional to your total flux by the surface. Consequently, if ɸ is the flux that is complete, E0 is electric constant, the total electric cost Q enclosed because of the surface is;

Q= ɸ E0

The formula is expressed by:

ɸ=Q/E0  

Where Q is the total charge inside the surface and E0 is the electric constant. 

Derivation 

Considering total flux through a sphere of radius, r encloses a point charge q at its centre. Divide the sphere into small area elements.

The flux through an area element ΔS is

Δɸ=E.ΔS=q4ℼE0r2r.ΔS

We used coulomb’s Law for the electric field due to a single charge q. Now, since the normal sphere at every point is along the radius vector at that point, the area element ΔS and rˆ have the same direction. Therefore, 

Δɸ=q4ℼE0r2ΔS

Since the magnitude of a unit vector is 1, the total flux through the sphere is obtained by adding up flux through all the different elements: 

ɸ= 𝝨q4ℼE0r2ΔS

Since each area element of the sphere are at the same distance r from the charge, 

ɸ= 𝝨q4ℼE0r2        ΔS=q4ℼE0r2S

Now S, the total area of the sphere, equal 4ℼr2

ɸ= q4ℼE0r2×4ℼr2=qE0

Gauss Law formula solved examples

Example 1: If the electric flux of a sphere is E×4ℼr2. What will be the electric field due to this flux?

Solution: The parameter is, 

ɸ=E×4ℼr2

Gauss formula: ɸ=Q/E0

Therefore, E×4ℼr2=Q/E0

E= Q(4ℼr2)E0

Example2: An electric flux of 2 V-m goes through a sphere in a vacuum space. What will be the charge that originates that flux? Use the Gauss law formula.

Solution: From the formula of Gauss law,

ɸ=Q/E0 

Q=ɸ×E0 

Substituting the value, we will get, 

Q=2V-m×8.85×10-12

Q=17.7×10-12C

Thus the charge to generate that flux will be 17.7×10-12C.

Example 3: A uniform area of the electric field of E=200N/c is present in the room in the X-direction. Utilising the Gauss theorem, calculate the flux of this area through a jet square of 10cm placed in the Y-Z plane. Take the normal along the positive Xaxis to be positive.

Solution: The flux ɸ= ∫E.cosϴ ds. 

A uniform electric field of magnitude E=200 N/C exists in the space along X-Axis

A plane square f edge 10cm is placed in the y-z plane 

When the plane is parallel to yz plane, angle b/w field and area ϴ=0

Also, E is uniform, 

So,

 ɸ=E.ΔS

ɸ=20010-2

ɸ=2 NC-1m2

Conclusion 

Gauss’ Law is a law applied to any closed surface. It is a significant apparatus since it allows the evaluation of the amount of enclosed charge by planning the field on a surface outside the charge distribution. According to the Gauss theorem, the total electric charge enclosed within a closed surface is proportional to your total flux by the surface. Consequently, if ɸ is the complete E0is electric constant flux, the total electric cost Q enclosed because of the surface is; Q= ɸ E0a.